Question

I've got to solve this initial value problem dy/dx = 2x / 3y^2-12 with y(-sqrt(65)) = 0

Answer 1

\[dy/dx = (2x)/(3y ^{2}-12)\]

Answer 2

Need to find C and then put it in implicit form

Answer 3

\[\frac{dy}{dx}=\frac{2x}{3y^2-12}\]
\[(3y^2-12)dy=2xdx\]
\[y^3-12y=x^2+C\]
\[y(-\sqrt{65})=0\quad\Rightarrow\quad C=-65\]
\[y^3-12y=x^2-65\]